Larger Summation

Note that $\displaystyle\sum_{x=1}^{\infty} \dfrac{1}{7x} = \dfrac{1}{7} \displaystyle\sum_{x=1}^{\infty} \dfrac{1}{x}$ is Harmonic series whose value doesn't converge since the value of sum depends upon the upper limit of sum .

Also the infinity sum is $1 < \displaystyle\sum_{x=1}^{\infty}\dfrac{1}{2^x -x} \leq 2$ Well I don't find any nice way to sum it less than 2. However, let's show it less than 2 $\begin{aligned} &{\color{#3D99F6}\dfrac{1}{2}} \phantom{c}< {\color{#D61F06}1} = {\color{#20A900}1} \\& {\color{#3D99F6}\dfrac{1}{2}+ \dfrac{1}{4}}\phantom{c} < {\color{#D61F06}1+ \dfrac{1}{2}} \geq{\color{#20A900} 1+\dfrac{1}{1+2}} \\& {\color{#3D99F6}\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}}\phantom{c} <{\color{#D61F06} 1 +\dfrac{1}{2}+\dfrac{1}{5}} \geq {\color{#20A900}1+\dfrac{1}{1+2} +\dfrac{1}{1+2+3}}\\& {\color{#3D99F6}\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16} + \cdots }< 1 + {\color{#D61F06}\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{12} +\cdots} \leq {\color{#20A900}1+ \dfrac{1}{1+2}+\dfrac{1}{1+2+3}+ \dfrac{1}{1+2+3+4}\cdots} \\& \displaystyle\sum_{x=1}^{\infty}\dfrac{1}{2^x} < \displaystyle\sum_{x=1}^{\infty} \dfrac{1}{2^x-x} \leq 2 \\& 1< \displaystyle\sum_{x=1}^{\infty} \dfrac{1}{2^x-x}\leq 2 \end{aligned}$

Therefore, $\small\displaystyle\sum_{x=1}^{\infty} \dfrac{1}{7x} \boxed{\large > }\displaystyle\sum_{x=1}^{\infty} \dfrac{1}{2^x-x}$ is the answer.

Relevant wiki: Convergence - Ratio Test

We note that the LHS $\displaystyle \sum_{x=1}^\infty \frac 1{7x} = \frac 17 \sum_{x=1}^\infty \frac 1x \to \infty$ . It is known that the harmonic series $\displaystyle H_n = \sum_{k=1}^n \frac 1k$ does not converge.

Now, let us apply the ratio test to the RHS:

$\begin{aligned} L & = \lim_{x\to \infty} \left| \frac {a_{x+1}}{a_x} \right| & \small \color{#3D99F6} \text{where }a_x \text{ is the }x \text{th term of the summation.} \\ & = \lim_{x\to \infty} \left| \frac {2^x-x}{2^{x+1}-x-1} \right| & \small \color{#3D99F6} \text{Divide up and down by }2^x \\ & = \lim_{x\to \infty} \left| \frac {1-\frac x{2^x}}{2-\frac x{2^x}-\frac 1{2^x}} \right| & \small \color{#3D99F6} \text{See note }\lim_{x\to \infty} \frac x{2^x} = 0 \\ & = \frac 12 < 1 \end{aligned}$

Implying that RHS converges and hence $\displaystyle \sum_{x=1}^\infty \frac 1{7x} \ \boxed{>} \ \sum_{x=1}^\infty \frac 1{2^x-x}$ .

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Note: reference: L'Hôpital's rule

$\begin{aligned} L & = \lim_{x \to \infty} \frac x{2^x} & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{x \to \infty} \frac 1{2^x\ln 2} & \small \color{#3D99F6} \text{Divide up and down w.r.t. }x \\ & = 0 \end{aligned}$